Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include<bits/stdc++.h>

using namespace std;
typedef long long LL;

const int MX=100005,MXT=1<<23|1;

int n,m,a[MX],b[MX],s[MX],L[MX],R[MX],x[MXT],l[MXT],r[MXT];
int p,q;
#define ll(i) (l[i]?l[i]:(l[i]=++q))
#define rr(i) (r[i]?r[i]:(r[i]=++q))
LL f[MX],ans;

inline void upd(int i,int li,int ri,int _l,int _r,int p){
	if(_l<=li&&ri<=_r)x[i]=p;
	else{
		if(x[i])x[ll(i)]=x[rr(i)]=x[i],x[i]=0;
		int mid=(li+ri)>>1;
		if(_l<=mid)upd(ll(i),li,mid,_l,_r,p);
		if(_r>mid)upd(rr(i),mid+1,ri,_l,_r,p);
	}
}
inline int que(int i,int li,int ri,int p){
	if(x[i])return x[i];
	if(!i||li==ri)return 0;
	int mid=(li+ri)>>1;
	return p<=mid?que(l[i],li,mid,p):que(r[i],mid+1,ri,p);
}
int main(){
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		cin>>a[i]>>b[i];
		if(b[i]==1&&a[i]*2>m){cout<<-1<<endl;return 0;}
		s[i]=(s[i-1]-2*a[i]%m+m)%m;
	}
	for(int i=1;i<=n;i++)
		if(b[i]==2)L[i]=1,R[i]=m;
		else L[i]=s[i-1]+1,R[i]=s[i]+1;
	for(p=1;p<m;p<<=1);q=1;
	for(int i=n,j;i>=2;i--){
		j=que(1,1,p,L[i]);
		if(j)f[i]=f[j]+(L[j]-L[i]+m)%m;
		if(L[i]<=R[i]){
			if(L[i]>1)upd(1,1,p,1,L[i]-1,i);
			if(R[i]<m)upd(1,1,p,R[i]+1,m,i);
		}else upd(1,1,p,R[i]+1,L[i]-1,i);
	}
	ans=1e18;
	for(int i=2,j,k;i<=2*n+1;i++){
		if(i&1)j=L[i>>1];
		else j=R[i>>1];
		if(L[1]<=R[1]&&L[1]<=j&&j<=R[1]||L[1]>R[1]&&(L[1]<=j||j<=R[1])){
			k=que(1,1,p,j);
			if(!k)ans=0;
			else ans=min(ans,f[k]+(L[k]-j+m)%m);
		}
	}
	for(int i=1;i<=n;i++)ans+=a[i]<<1;
	cout<<ans<<endl;
	return 0;
}